Question

十字相乘法分解因式

（1）x^2+5x+8 （2）2x^2-7x+3 (2)-2x^2-3x+2 (3)-a^2+10a-9
(4)5x^2+7xy-6y^2 (5)-2(a+b)^2+(a+b)+3 (6)(x+y)^4+4(x+y)^2-5
x^3-7x^2+10x

Answer 1

（1）x^2+6x+8
=(x+2)(x+4)

（2）2x^2-7x+3
=(2x-1)(x-3)

(2)-2x^2-3x+2
=-(2x²+3x-2)
=-(2x-1)(x+2)

(3)-a^2+10a-9
=-(a²-10a+9)
=-(a-1)(a-9)

(4)5x^2+7xy-6y^2
=(5x-3y)(x+2y)

(5)-2(a+b)^2+(a+b)+3
=-[2(a+b)²-(a+b)-3]
=-(2a+2b-3)(a+b+1)

(6)(x+y)^4+4(x+y)^2-5
=[(x+y)² +5][(x+y)²-1]
=(x²+2xy+y²+5)(x+y-1)(x+y+1)

x^3-7x^2+10x
=x(x²-7x+10)
=x(x-2)(x-5)

Answer 2

（1）x^2+6x+8
=(x+2)(x+4)

（2）2x^2-7x+3
=(2x-1)(x-3)

(2)-2x^2-3x+2
=-(2x²+3x-2)
=-(2x-1)(x+2)

(3)-a^2+10a-9
=-(a²-10a+9)
=-(a-1)(a-9)

(4)5x^2+7xy-6y^2
=(5x-3y)(x+2y)

(5)-2(a+b)^2+(a+b)+3
=-[2(a+b)²-(a+b)-3]
=-(2a+2b-3)(a+b+1)

(6)(x+y)^4+4(x+y)^2-5
=[(x+y)² +5][(x+y)²-1]
=(x²+2xy+y²+5)(x+y-1)(x+y+1)

x^3-7x^2+10x
=x(x²-7x+10)
=x(x-2)(x-5)

Answer 3

（2）2x^2-7x+3
2x 1
x 3
（2x-1）（x-3）
-2x^2-3x+2
-2x 1
x 2
-a^2+10a-9
-a 1
a -9
5x^2+7xy-6y^2
5x -3y
x 2y
(5)-2(a+b)^2+(a+b)+3
-2(a+b) 3
(a+b) 1
(6)(x+y)^4+4(x+y)^2-5
(x+y)^2 5
(x+y)^2 -1
x^3-7x^2+10x=x(x^2-7x+10)
x -5
x -2

Answer 4